博耶表达 [英] boyer expression
问题描述
请知道下面这个表达方式的人与我分享一些信息..
（x2 = x1 + s）& （y2 = y1）& （f（x1，y1）> -1）=> f（x1，y1）= f（x2，y2）[/ i]
这种关系是什么意思？左边的expr和右边是否相等？
提前谢谢..
请知道以下表达方式的人与我分享一些信息..
（x2 = x1 + s）& （y2 = y1）& （f（x1，y1）> -1）=> f（x1，y1）= f（x2，y2）[/ i]
这种关系是什么意思？左边的expr和右边是否相等？
提前谢谢..
它只是一个条件关系：A => ; B读作如果A为真，那么B为真。
在你的例子中，A是（x2 = x1 + s）& （y2 = y1）& f（x1，y1）> -1）"
读作三和子句：a和b和c。类似的推理适用
箭头右侧的表达式B.
整个表达式取决于''s'的值和当然，定义
函数''f''。在不知道这些定义的情况下，整个关系是无意义的。
亲切的问候，
Jos
它只是一个条件关系：A => B读作如果A为真，那么B为真。
在你的例子中，A是（x2 = x1 + s）& （y2 = y1）& f（x1，y1）> -1）"
读作三和子句：a和b和c。类似的推理适用
箭头右侧的表达式B.
整个表达式取决于''s'的值和定义
函数''f'当然。在不知道这些定义的情况下，整个关系是无意义的。
亲切的问候，
Jos
非常感谢Jos ...第一次看expr我真的很无能。
当你把''如果A是真的话那么B是真的''它'真的很容易理解。
所以，''f''这里可能是任何计算函数，如除法（/）或字符串搜索算法 - boyer？
最后一个expr需要咨询你...
（x3 = x1 + x2）& （y1 = y2 = y3）& （f（x，1，y1）= -1）=> f（x3，y3）< = f（x2，y2）+ len（x1）
根据我的理解，该关系显示了与字符串x1的长度的关系。
其他任何方式.....？再次感谢你...
非常感谢Jos ...第一次看看expr我真的很无能为力。 />
当你把''如果A是真的那么B是真的''它真的很容易理解。
所以，''f''这里可以是任何计算分裂（/）或字符串搜索算法等函数 - boyer？
最后一个expr需要咨询你...
（x3 = x1 + x2）& （y1 = y2 = y3）& （f（x，1，y1）= -1）=> f（x3，y3）< = f（x2，y2）+ len（x1）
根据我的理解，该关系显示了与字符串x1的长度的关系。
其他任何方式.....？再次感谢你...
你在读什么文章？我敢肯定，这些darn公式周围有一些解释文字。我的猜测确实是x1，x2，x3和y1，y2和y3
是字符串，'+''是字符串连接运算符，我认为（只是一个猜测）
''f''是Boyer Moore字符串匹配算法中使用的失败函数。
亲切的问候，
Jos
Kindly please someone who knows about this below expression share some info with me..
(x2 = x1 + s) & ( y2 = y1) & (f (x1, y1) > -1) => f(x1, y1) = f(x2, y2)[/i]
What does the relation mean? Are the left expr and right equal?
Thank you in advance..
Kindly please someone who knows about this below expression share some info with me..
(x2 = x1 + s) & ( y2 = y1) & (f (x1, y1) > -1) => f(x1, y1) = f(x2, y2)[/i]
What does the relation mean? Are the left expr and right equal?
Thank you in advance..It''s just a conditional relation: A => B reads as "if A is true then B is true".
In your example A is "(x2 = x1+s) & (y2 = y1) & f(x1, y1) > -1)" which
reads as three and clauses: "a and b and c". Similar reasoning applies
the the expression B at the right side of the arrow.
The entire expression depends on the value of ''s'' and the definition of the
function ''f'' of course. Without know those definitions the entire relation is
meaningless.
kind regards,
Jos
It''s just a conditional relation: A => B reads as "if A is true then B is true".
In your example A is "(x2 = x1+s) & (y2 = y1) & f(x1, y1) > -1)" which
reads as three and clauses: "a and b and c". Similar reasoning applies
the the expression B at the right side of the arrow.
The entire expression depends on the value of ''s'' and the definition of the
function ''f'' of course. Without know those definitions the entire relation is
meaningless.
kind regards,
JosThank you very much Jos...first time look at the expr I was really clueless.
When you put ''if A is true then B is true'' it''s really simple to understand.
So, ''f'' here could be any calculation functions such as division (/) or string search algorithm - boyer?
One last expr need to consult you...
(x3 = x1 + x2) & (y1 = y2 = y3) & (f (x,1, y1) = -1) => f(x3, y3) <= f(x2,y2) +len(x1)
From my understanding, the relation shows relation with lenght of string x1.
Any other means .....? Thank you again...
Thank you very much Jos...first time look at the expr I was really clueless.
When you put ''if A is true then B is true'' it''s really simple to understand.
So, ''f'' here could be any calculation functions such as division (/) or string search algorithm - boyer?
One last expr need to consult you...
(x3 = x1 + x2) & (y1 = y2 = y3) & (f (x,1, y1) = -1) => f(x3, y3) <= f(x2,y2) +len(x1)
From my understanding, the relation shows relation with lenght of string x1.
Any other means .....? Thank you again...What article are you reading? I''m sure that there''s some explaining text around
those darn formulas. My guess is that indeed x1, x2, x3 and y1, y2 and y3
are strings, ''+'' is the string concatenation operator and I think (just a guess) that
''f'' is the failure function used in the Boyer Moore string matching algorithm.
kind regards,
Jos
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